find the domain of the function f(x)= 1/sqrt (4x-8)
please show calculations

- anonymous

find the domain of the function f(x)= 1/sqrt (4x-8)
please show calculations

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

huh

- anonymous

callum can you show the steps please

- anonymous

(2, infinity)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

\[f(x) = \frac{1}{\sqrt{4x-8}}\] now you want to look at conditions on 4x-8. What can x reasonably be? Well you cannot take a square root of a negative number, and the denominator cannot be zero, so we need:
\[4x-8 > 0\]
\[x>\frac{1}{2}\]
So I was wrong above, the domain is actually:
\[\left(\frac{1}{2}, \infty \right)\]

- anonymous

x > 2

- anonymous

think you divided the wrong side there

- anonymous

oh yes sorry it's 2 lol, quite right.
\[(2,\infty)\]

- anonymous

rickjbr can you demonstrate how you got your answer please

- anonymous

you'll have imaginary numbers if the square root is allowed to go negative

- anonymous

thus not part of the real number system

- anonymous

callum was right in his calculation
4x - 8 > 0
4(x-2) > 0
x-2 > 0
x > 2

- anonymous

yeah my reasoning was right, it's just I divided the inequality wrong and got x>1/2. rickjbr corrected this to x>2

- anonymous

thanks for the help to both of your

- anonymous

np

- anonymous

how about this. 2x^2-3x+5...find and simplify the difference quotient

- anonymous

The question reads
Find and simplify the difference quotient f(x+h)-f(x)/h for the given function f(x)=2x^(2)-3x+5

- anonymous

ah, right

- anonymous

(2(x+h)^2-3(x+h)+5-2x^2+3x-5)/h

- anonymous

(2(x^2+2xh+h^2)-3(x+h)+5-2x^2+3x-5)/h

- anonymous

(2x^2+4xh+2h^2-3x-3h+5-2x^2+3x-5)/h

- anonymous

(2h^2+4xh+3x)/h

- anonymous

is that the final answer

- anonymous

checking, a bit rusty

- anonymous

my options are a)4x+2h, b)4x+2h-3, c)4h+2x-3, d)4h+2x, e)4x+4h

- anonymous

i'll redo it, might've missed something

- anonymous

ok

- anonymous

ah, silly

- anonymous

that last step was it. instead of (2h^2+4xh+3x)/h
it should be (2h^2+4xh-3h)/h

- anonymous

simply factor out an h

- anonymous

[h(2h+4x-3)]/h
=2h+4x-3

- anonymous

that's it

- anonymous

thanks greatly. i dont know if you have the time...but im reviewing for a test and I have a whole lot of other questions. if you can't help me, i understand, i will just post on the main forum

- anonymous

probably better off posting on main forum, if i can help, i'll try

- anonymous

ok thanks for the help

- anonymous

Find \[\frac{f(x+h)-f(x)}{h}\] for \[f(x)=2x^2-3x+5\]
\[\frac{f(x+h)-f(x)}{h} = \frac{(2(x+h)^2-3(x+h)+5) - (2x^2-3x+5)}{h}\]
\[= \frac{(2x^2+4xh+2h^2-3x-3h)-(2x^2-3x)}{h}\]
\[=\frac{4xh+2h^2-3h}{h} = 4x+2h-3\]
We can verify that this is right by noting that
\[f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}\]
The left hand side of this is equal to the derivative of f, which we can just do right here:
\[f'(x) = 4x - 3\]
and we should get that this equals the limit as h goes to zero of \[\frac{f(x+h) - f(x)}{h}.\] We have already calculated that \[\frac{f(x+h) - f(x)}{h} = 4x+2h-3\] so taking the limit of this as h goes to zero we get:
\[\lim_{h\rightarrow 0}(4x+2h - 3) = 4x+0-3 = 4x-3.\]
Hence we ave verified that the simplification has been done correctly.

Looking for something else?

Not the answer you are looking for? Search for more explanations.